Sedimentation coefficients, using a centrifuge, are expressed using Svedberg unit (symbol S, sometimes Sv). Wikipedia states that $S = 10^{-13}$ sec but I also saw in a book that actually $S = 10^{-13}$ sec/$\text{rad}^2$. Which of these is correct?

I know that a 50S ribosome has higher molecular weight than a 30S ribosome and for that reason the 50S would travel faster (on a test tube placed in the centrifuge) than a 30S, but if substitute S by $10^{-13}$ sec, for me it's a little bit-non intuitive, because 50S would be $50\cdot{10^-13}$ sec (and that is greater than $30\cdot{10^-13}$ sec), which means it takes more time to travel? Obviously not, so what's the physical meaning of time in this situation?

In response to Alan Boyd:

"In other words, at 1x g a 1S particle will travel at $10^{-13} m\cdot s^{-1}$" or $10^{-12}$? Indeed 5.5h is a reasonable value but I still cannot understand the significance of the ratio between velocity and acceleration which corresponds to Svedberg units (in seconds). Maybe $50\cdot{10^-13}$ sec is the time that our ribosome takes to reach the terminal velocity in the fluid? That would make some sense, for example, if I drop an unfolded A4 paper to the ground the time it will take to reach terminal velocity will be lower than the time a A4 folded paper will take to reach it's specific terminal velocity, which is higher than the velocity from the unfolded A4 paper, because the area is larger (the friction force acting on it will be very high) which is consistent with the observations that are made (in centrifugation objects with larger surface area will travel at a slower terminal velocity). What do you think ?


1 Answer 1



The sedimentation coefficient is the ratio of the speed of a substance in a centrifuge to its acceleration in comparable units. A substance with a sedimentation coefficient of 26S (26×10−13 s) will travel at 26 micrometers per second (26×10−6 m/s) under the influence of an acceleration of a million gravities (107 m/s2). Centrifugal acceleration is given as rω2; where r is the radial distance from the rotation axis and ω is the angular velocity in radians per second.

In other words, at 1x g a 1S particle will travel at 10-13 m s-1 = 0.1 pm s-1

Particles with higher values of S will travel proportionately faster, and increasing g force will also increase sedimentation rate.

Let's look at the example of a 50S ribosome at 100 000 g:

rate of sedimentation = 10-13 x 105 x 50 m per sec = 50 x 10-8 m s-1

So how long to travel 10 cm = 10-2 m?

=10-2/(50 x 1010-8) s

= approximately 5.5 h, which seems about right.

  • $\begingroup$ I edited my question, in response to your answer because my comment was too large. $\endgroup$
    – bruno
    May 17, 2014 at 15:43
  • $\begingroup$ You are overthinking. Our ribosome is in circular motion so is experiencing a constant acceleration, which is why the WP quotation goes on to define centrifugal acceleration. en.wikipedia.org/wiki/Circular_motion $\endgroup$
    – Alan Boyd
    May 17, 2014 at 16:32
  • $\begingroup$ I'm asking for the intuition to compare the value of 50S and 30S in seconds. I can go to the formula and say that the centriptal acceleration experienced from both objects is the same therefore to have a greater S the terminal velocity must be greater and obviously 50S expirience a greater terminal velocity, but what does that quotient mean? What does the time indicate? $\endgroup$
    – bruno
    May 17, 2014 at 16:58

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